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Electronics and Communication Engineering (ECE) Exam  >  Electronics and Communication Engineering (ECE) Tests  >  RRB JE Mock Test Series for ECE 2025  >  RRB JE ECE (CBT I) Mock Test- 4 - Electronics and Communication Engineering (ECE) MCQ

RRB JE ECE (CBT I) Mock Test- 4 - Electronics and Communication Engineering (ECE) MCQ


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30 Questions MCQ Test RRB JE Mock Test Series for ECE 2025 - RRB JE ECE (CBT I) Mock Test- 4

RRB JE ECE (CBT I) Mock Test- 4 for Electronics and Communication Engineering (ECE) 2024 is part of RRB JE Mock Test Series for ECE 2025 preparation. The RRB JE ECE (CBT I) Mock Test- 4 questions and answers have been prepared according to the Electronics and Communication Engineering (ECE) exam syllabus.The RRB JE ECE (CBT I) Mock Test- 4 MCQs are made for Electronics and Communication Engineering (ECE) 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for RRB JE ECE (CBT I) Mock Test- 4 below.
Solutions of RRB JE ECE (CBT I) Mock Test- 4 questions in English are available as part of our RRB JE Mock Test Series for ECE 2025 for Electronics and Communication Engineering (ECE) & RRB JE ECE (CBT I) Mock Test- 4 solutions in Hindi for RRB JE Mock Test Series for ECE 2025 course. Download more important topics, notes, lectures and mock test series for Electronics and Communication Engineering (ECE) Exam by signing up for free. Attempt RRB JE ECE (CBT I) Mock Test- 4 | 100 questions in 90 minutes | Mock test for Electronics and Communication Engineering (ECE) preparation | Free important questions MCQ to study RRB JE Mock Test Series for ECE 2025 for Electronics and Communication Engineering (ECE) Exam | Download free PDF with solutions
RRB JE ECE (CBT I) Mock Test- 4 - Question 1
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If sinθ + sin2θ = 1, then the value of cos12θ + 3 cos10θ + 3 cos8θ + cos6θ – 1 is:

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 1
sinθ + sin2θ = 1 ⇒ sinθ = cos2θ

Given : cos12θ + 3 cos10θ + 3 cos8θ + cos6θ–1

= (cos4θ + cos2θ)3 – 1 = (sin2θ + cos2θ)3 – 1 = 1 – 1 = 0

RRB JE ECE (CBT I) Mock Test- 4 - Question 2
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The height of an equilateral triangle is 4√3 cm, then what will be the area of the equilateral triangle?

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 2

Height = 4 cm

Area of equilateral triangle = √3/4 x 82 = 16√3 cm2.

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RRB JE ECE (CBT I) Mock Test- 4 - Question 3
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If 10 men can do work in 6 days and 15 women can do the same in 5 days, Then 8 men and 5 women can together do the work in :

Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 3
10 Men can do the work in 6 days

If the work is to be finished in 5 days, number of women required = 15

If the work is to be finished in 6 days, number of women required =[15×5]/6 = 25/2

So that 10 men = 25/2 women

or 5 women = 4 men

8 men + 5 women = 8 men + 4 men = 12 men

Now 10 men can finish the work in = 6 days

∴ 12 men can finish the work in [6×10]/12=5 days

RRB JE ECE (CBT I) Mock Test- 4 - Question 4
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The average of two numbers is 8, and that of another three numbers is 3. The average of these five numbers is-
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 4
Sum of two numbers = 8 ×2 = 16

Sum of other three numbers = 3 × 3 = 9

Total sum = 25

Average = 25/5 = 5

RRB JE ECE (CBT I) Mock Test- 4 - Question 5
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Which of the following is obtained after the rationalization of the expression 1/(√2+√3+√5) ?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 5

RRB JE ECE (CBT I) Mock Test- 4 - Question 6
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If sin θ=8/17, where 0o<θ<90o, then tan θ+sec θ is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 6
Using pythagoras theorem

AC2 = AB2 + BC2

(17)2 = AB2 + (8)2

AB2 = 289 - 64

AB = √225

AB = 15

∴ tan θ = 8/15 and sec θ = 17/15

∴ tan θ + sec θ = 8/15 + 17/15 = 25/15 = 5/3

RRB JE ECE (CBT I) Mock Test- 4 - Question 7
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OA, OB, OC are 3 lines in a plane which meet at O. If the angles ∠AOB, ∠BOC, ∠COA are 2x, 5x, 8x respectively (where x is a positive angle), then x equals:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 7

2x + 5x + 8x = 360°

15x + 360°

x = 360°/15 = 24°

RRB JE ECE (CBT I) Mock Test- 4 - Question 8
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The value of tan 31° tan 33° ... tan 59° is equal to:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 8

tan 31°. tan 32°. tan 33° .... tan 59°

= tan 31°. tan 32°. tan 33° ... tan 57°. tan 58°. tan 59°

= tan 31º. tan 32°. tan 33° ... cot 33°. cot 32°. cot 31° = 1

[tan (90°-θ) = cotθ and tanθ cotθ =1]

RRB JE ECE (CBT I) Mock Test- 4 - Question 9
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If x = 81, then the value of (x1/4 -1) (x1/4+1) is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 9

x1/2 - 1 = (81)1/2 - 1 = 9 - 1 = 8

RRB JE ECE (CBT I) Mock Test- 4 - Question 10
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A man sells two scooters for Rs.12000 each. He makes a profit of 20% on one and a loss of 20% on the other. The profit/loss, on the whole, is
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 10

SP1 = 12000

profit = 20%

∴ CP1 = 12000/120 x 100 = Rs.10,000

SP2 = 12000

Loss % = 20%

∴ CP = 12000/80 x 100 = Rs.15,000

Total CP = 25000

Total SP = 24000

Loss = Rs.25000 – Rs.24000 = Rs.1000

(or)

If the selling price of the two items is equal and the percentage of profit on one equals the percentage of loss on the other, then, the result is always loss and the percentage of loss is (loss%2/100).

Hence, in the present case, loss percentage = (20)2/100 percent = 4%.

Loss = 4/96 × Rs.24000 = Rs.1000

RRB JE ECE (CBT I) Mock Test- 4 - Question 11
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A train covers a certain distance in 50 minutes, if it runs at a speed of 48 km/hr on an average. The speed at which the train must run so that the time of journey becomes 40 minutes?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 11
We are having time and speed given, so first we will calculate the distance. Then we can get a new speed for a given time and distance.

Time = 50/60 hr = 5/6 hr

Speed = 48 mph

Distance = S*T = 48 * 5/6 = 40 km

New time will be 40 minutes so,

Time = 40/60 hr = 2/3 hr

Now we know,

Speed = Distance/Time

New speed = 40*3/2 kmph = 60kmph

RRB JE ECE (CBT I) Mock Test- 4 - Question 12
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The pie chart given here shows expenditures incurred by a family on various items and their savings, which amount to Rs. 8,000 in a month.

Study the chart and answer the questions based on the pie chart.

How much expenditure is incurred on education?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 12

∴ 60° = Rs8000

∴ Total expenditure

300° = 8000/60 x 300 = rs 40,000

Expenditure on education

= 30/300 x 40,000 = rs 4000

RRB JE ECE (CBT I) Mock Test- 4 - Question 13
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The pie chart given here shows expenditures incurred by a family on various items and their savings, which amount to Rs. 8,000 in a month.

Study the chart and answer the questions based on the pie chart.

The ratio of the expenditure on food to the savings is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 13
Expenditure on food

=(120/300)×40,000=₹16,000

Required ratio = 16000 : 8000 ⇒ 2 : 1

RRB JE ECE (CBT I) Mock Test- 4 - Question 14
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A solid metallic cone of height 10 cm and radius of base 20 cm is melted to make spherical balls each of 4 cm diameter. How many such balls can be made?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 14
Volume of cone = 1/3πr2h

= (1/3)×π×(20)2×10

= 400×10

= (π/3)×4000=(4000/3)π

Volume of spherical balls = 4/3πr3

= (4/3)×π×(2)2

= 32/3π

∴ Spherical balls are formed from the cones.

∴ Total volume of all spherical cones = Volume of cones.

∴ No. of spherical balls = Volume of cone/Vol of 1 spherical ball

= 4000/3

= 32/3

= 4000/32=125

RRB JE ECE (CBT I) Mock Test- 4 - Question 15
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Sum of the lengths of any two sides of a triangle is always greater than
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 15
The sum of the other two sides of a triangle is always greater than the third side.
RRB JE ECE (CBT I) Mock Test- 4 - Question 16
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If D is a point on the side AB of triangle ABC and DE is a line through D meeting AC at E such that ∠ADE = ∠ACB, then AB. AD is equal to:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 16
Clearly ΔADE and ΔABC are similar

∴ AD/AC = AE/AB

AD. AB = AE. AC

RRB JE ECE (CBT I) Mock Test- 4 - Question 17
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A man can row 15 km per hour downstream and 9 km per hour upstream. The speed (in km/hour) of the boat in still water is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 17

The Speed of Boat in still water = 1/2 (Rate downstream + Rate upstream)

The speed of the boat in still water = 1/2 (15 +9) = 24/2 = 12 km/hr.

RRB JE ECE (CBT I) Mock Test- 4 - Question 18
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If x : y = 3 : 4, then the value of (5x - 2y)/(7x+2y)
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 18

RRB JE ECE (CBT I) Mock Test- 4 - Question 19
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The sum of the numerator and denominator of a certain fraction is 8. If 2 is added to both the numerator and the denominator the value of the fraction increased by 4/35, then the fraction is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 19

x + y = 8 ……(i)

or 35 (y - x) = 2y (y + 2) (ii)

from (i)

x = 8 - y

substituting this value in (ii)

35 (y - 8 + y) - 2y (y + 2)

35 (y - 4) = y (y + 2)

y2 - 33y + 140 = 0

y = 28 or 5

We can't take 28, taking y = 5, x will become 3

Therefore x/y = 3/5

RRB JE ECE (CBT I) Mock Test- 4 - Question 20
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The simple interest on a certain principal at the rate of 9.5 percent per annum is Rs 950 for two year. How much will be the additional interest on the same amount for the same period at the rate of 10.5 percent per annum?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 20

S.I. = Rs 950, rate = 9.5%

time = 2 years

∴ sum = (950×100)/(9.5×2)=5000

If rate of interest is 10.5% then

Simple Interest = (5000×2×10.5)/100 = Rs 1050

∴ additional interest due to change in rate of interest = Rs 100

RRB JE ECE (CBT I) Mock Test- 4 - Question 21
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If the angle of elevation of the top of a tower from two points at the distance x and y meter from the base and in the same straight line with it are complementary, then the height of the tower is?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 21

Let AB be the tower and C and D be the point of observation on AC.

∠AC B = θ, ∠ADB = 90 - θ, AB = h m

AC = X m, AD = Y m

So, CD = x - y m.

tan θ = h/x

And tan (90 - θ) = h/y

RRB JE ECE (CBT I) Mock Test- 4 - Question 22
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Kamya purchased an item of 46,000 and sold it at a loss of 12%. With that amount she purchased another item and sold it at a gain of 12%. What was her overall gain/loss?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 22
SP of first article = (46000×88)100=₹40,480

SP for second article = (40480×112)/100=₹45,337.60

∴ Loss = 46000 - 45337.60 = ₹662.40.

RRB JE ECE (CBT I) Mock Test- 4 - Question 23
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The owner of a cell phone shop charges his customer 23% more than the cost price. If a customer paid Rs.7,011 for a cell phone, then what was the cost price of the cell phone?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 23
Let the cost price of cell phone be Rs. x

According to the question,

[(100+23)/100] × x = 7011

Or, x = (7011×100)/123 = 5700

RRB JE ECE (CBT I) Mock Test- 4 - Question 24
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The cost of an article was Rs.75. The cost was first increased by 20% and later on it was reduced by 20%. The present cost of the article is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 24
Effective decrease

= [20 – 20 – 20×(20/100)] % = -4%

∴ Present cost of the article = 96% of Rs.75

= (75×96)/100 = Rs.72

RRB JE ECE (CBT I) Mock Test- 4 - Question 25
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A rectangular box measures internally 1.6 m long, 1 m broad and 60 cm deep. The number of cubical block each of edge 20 cm that can be packed inside the box is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 25
Volume of rectangular box = 160 x 100 x 60 cm3

and volume of one cubical block = 20 x 20 x 20 cm3

∴ Required number of cubical block = (160×100×60)/(20×20×20) = 120 blocks

RRB JE ECE (CBT I) Mock Test- 4 - Question 26
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The diameter of the curved surface of a bucket is 28 decimeter and 14 decimeters and its height is 12 decimeter. Find its volume.
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 26

RRB JE ECE (CBT I) Mock Test- 4 - Question 27
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What will be the area of a rhombus, whose one side is 20 cm and one diagonal is 24 cm?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 27
Let length of another diagonal = 2x

Diagonals of rhombus intersect at right angles, so by using pythagoras theorem,

x2 + 122 = 202

Therefore x = 16 , So diagonal = 2'16 =32

Area of rhombus

RRB JE ECE (CBT I) Mock Test- 4 - Question 28
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The ratio of two unequal sides of a rectangle is 1 : 2. If its perimeter is 24 cm, then the length of diagonal in cm is:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 28
Let the sides of rectangle be 2x, x cm

∴ Perimeter of rectangle = 2 (2x+x)

2 (2x+x) = 24

or 6x = 24

∴ x = 4 cm

hence length of diagonal = √[(2x)2+x2]

= √5x2

= √(5×16)

=√ 80

= 4√5

RRB JE ECE (CBT I) Mock Test- 4 - Question 29
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8 hours of work for a woman is equal to 6 hours of work for a man or 12 hours of work for a boy. If 9 men working 6 hours daily can complete a work in 6 days, then in how many days 12 men, 12 women and 12 boys working 8 hours daily will complete the same work?
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 29
1 man done work in 6×6×9 hours

Since 8W=6M=12B

therefore 12M+12W+12B=27M

therefore 27 men will do work in (6×6×9/27) = 12 hours

= 12/8=1.5 days

RRB JE ECE (CBT I) Mock Test- 4 - Question 30
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In an examination, the average marks was found to be 50. For deducting marks for computational errors, the marks of 100 candidates had to be changed from 90 to 60 each and so the average of marks came down to 45. The total number of candidates, who appeared at the examination, was:
Detailed Solution for RRB JE ECE (CBT I) Mock Test- 4 - Question 30
Let the total number of candidates be x.

∴ 50x - 30 × 100 = 45x

⇒ 5x = 3000

⇒ x = 3000/5 = 600

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